Designing a reinforced concrete bridge for a railway. Design of a road reinforced concrete bridge

Petersburg State University

Ways of Communication.

Department of "Bridges".

Skorik O.G.

Course project "Reinforced concrete bridge"

Explanatory note

Head: Completed:

Skorik O.G. Zholobov M.I.

St. Petersburg.

Part 1. Development of a variant ……………………………………… ... 3-6

Part 2. Calculation of the beam superstructure ……….…. …… ... 7-22

2.1. Calculation of the carriageway of span structures ………………… ..7-13

2.1.1. Determination of the calculated efforts ............................................................. 7-8

2.1.2. Calculation of slab sections ……………………………………… .... 8-13

2.2. Calculation of the main beams of the superstructure ………………… .13-23

2.2.1. Determination of the design forces .............................................................................. 13-14

2.2.2. Calculation of a beam from prestressed reinforced concrete …………………………………………………………………… .14-22

Part 3. Calculation of the intermediate support …………………. ……… ..23-27

3.1. Determination of the design forces in the elements of the supports ...................... 23-24

3.2. Calculation of sections of concrete supports …………………… ... ……… ... 24-27

References …………………………………………………… .28

Part 1. Development of a variant.

Appointment of basic dimensions.

The total length of the bridge is determined by a given bridge opening, taking into account the number of spans in the bridge scheme and the structural parameters of the supports (type of abutment, thickness of the intermediate support, etc.).

The required length of the bridge with loose abutments is calculated by the formula:

L p = l 0 + n * b + 3 * H + 2 * a, where

L p - the required length of the bridge between the ends of the abutments, m;

N is the number of intermediate supports falling into the water, m;

B is the average thickness of the intermediate support, m;

H-height from the middle line of the trapezoid formed by the horizontals of high and low-water waters (along which the opening of the bridge is measured) to the edge of the canvas, m;

L 0 - bridge hole, m;

A-value of the abutment entry into the embankment

(a = 0.75 at<6м. и a=1 при высоте насыпи>6m).

Thus

L p = 65 + 2 * 3.5 + 3 * 6.95 + 2 * 1 = 94.85m.

PR = DCS + h page + h gab = 22 + 2.75 + 5 = 29.75m.

BP = PR-0.9 = 29.75-0.9 = 28.85m.

H = 28.85- (23 + 20.8) * 0.5 = 6.95m.

Pile abutments are accepted. The length of the abutment wing on top with the span of the adjoining beams of 16.5 m will be 3.75 m. The actual length of the bridge with the adopted structures will be (taking into account the distance between the ends of the beams by 0.05):

L f = 3.75 + 0.05 + 16.5 + 0.05 + 27.6 + 0.05 + 27.6 + 0.05 + 16.5 + 0.05 + 3.75 =

The actual length of the bridge exceeds the full design

0.01 or 1%, which is permissible by the norms.

Determination of the scope of work

Spans. The volume of the reinforced concrete of the superstructure with the total length of 27.6 m is 83.0 m 3. The volume of the reinforced concrete of the superstructure with the total length of 16.5 m is 35.21 m 3.

Intermediate supports. We have three intermediate supports with a height of 5.3 m. The volume of reinforced concrete blocks is for one support:

V bl = 30.3m 3

Block grouting concrete and support filling concrete is

V ohm = m 3.

The volume of the grillage with a height of 2 m from monolithic reinforced concrete with dimensions in terms of 8.6 * 3.6 m with bevels of 0.5 m:

V height. = 2 * (3.6 * 8.6-4 * 0.5 3) = 60.92 m 3.

When assigning the dimensions of the intermediate supports, it is necessary to take into account the requirements of the norms, which indicate how the dimensions of the sub-truss plates of the intermediate supports are determined.

Based on the presence of ice drift, we arrange a rounded support. For a slab with a rounded shape in plan, the width and thickness are determined by the formulas:

a = e + c 1 + 0.4 + 2k 1;

b = m + c 2 + 0.4 + 2k 2;

Based on the tabular data, we get the following values:

a = 0.75 + 0.72 + 0.4 + 2 * 0.15 = 2.17m;

b = 1.8 + 0.81 + 0.4 + 2 * 0.3 = 3.61m;

An approximate calculation method can be used to determine the number of piles in the pile foundation of the intermediate support of a girder bridge.

The number of piles is determined by the formula:

n = m , where

M-coefficient, taking into account the influence of the bending moment acting on the base of the grillage, equal to 1.5-1.8;

SN is the sum of the calculated vertical forces acting on the base of the foundation.

SN = N bp + N ball + N pr. P. + N op.

Here N time, N ball, N pr. Page. , N op vertical pressures, tf, respectively, from the temporary load when loading two adjacent spans, from the weight of ballast on the spans of a railway bridge, from the weight of reinforced concrete spans and from the weight of the support with the foundation.

The indicated values ​​are determined by the formulas

N BP = g * to e;

N ball = 2.0 * 1.3 * F b *;

N pr. Page = 1.1 * V pr. Page * 2.5 * 0.5;

N op = 1.1 * V op * 2.4, where

L 1, l 2 - full lengths of spans, supported by supports, m;

G-safety factor for temporary load;

2.0-volumetric mass of ballast;

1.3-safety factor for ballast;

F b - cross-sectional area for ballast trough, m 2;

1,1 is the safety factor for the dead weight of the structure;

V pr.str - the volume of reinforced concrete spans resting on the support;

2.5-volumetric weight of reinforced concrete, t / m 3

V op - the volume of the body of the support and foundation, m 3;

P d - design bearing capacity of one pile (shell pile);

N BP = 1.2 * 14 * = 463.68 tf.

N ball = 2 * 1.3 * 1.8 * = 129.17 tf.

N pr.str = 1.1 * 2.5 * 0.5 * (83.0 + 83.0) = 228.25 tf.

N op = 1.1 * 2.4 * (61.42 + 30.3 + 46.51) = 364.93 tf.

åN = 458.05 + 129.17 + 228.25 + 364.93 = 1180.4 tf.

When using piles with a diameter of 60 cm 2 and a length of 15 m, the bearing capacity of the pile on the ground will be 125 tf and then the required number of piles

n = 1.6 * m.

Let's take 15 piles with a diameter of 60cm and a length of 15m for a support. The volume of hollow piles with a wall thickness of 8 cm will be

V ps = 15 * 15 * ( ) = 29.4m 3.

Volume of concrete for filling hollow piles

V s = 15 * 15 * m 3.

The pit fencing is made of a cobbled wooden sheet pile with a tongue length of 6 m, with a perimeter of the fence 2 * (5.6 + 10.6) = 32.4 m, the area of ​​the vertical walls will be 6 * 32.4 = 194.4 m 2.

Stay. The volume of reinforced concrete of the abutment head is 61.4 m 3

The volume of 9 hollow piles with a wall thickness of 8 cm and a length of 20 m will be

20*9*() = 24.1m 3.

The volume of concrete for filling the hollow piles of the abutment

20*9*27.4 m 3;

The scope of work and the determination of the costs of structural elements of the bridge are given in the table. Table 1

Reinforced concrete bridge design. Determination of the number of bridge spans. Bridge layout. Designing a bridge option for given local conditions is a multi-tasking task. possible solutions from which you must choose the best.

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1. V maintenance ……………………………………………. ……………………….… 2

2. Design of a reinforced concrete bridge…. ……………………………….… 3

3. Intermediate pore scheme …………………… .. ……………… .. ……… ...... 4

4. Determination of the number of piles in the foundation of the support…. …………………… ... …… .7

5. Determination of the number of bridge spans …………………………………… ...... 12

6.Scheme of the bridge ………………………………………………………………… ..14

7. References ……… ... ……………………………………………… ..15

INTRODUCTION

Designing a bridge option for given local conditions is a challenge with many possible solutions, from which it is necessary to choose the best. The complexity of solving this problem is associated, on the one hand, with a wide variety of systems and structures of reinforced concrete bridges, and as a consequence, a large number of bridge options that can be assigned to each bridge crossing. On the other hand, as a rule, it is not easy to find among the options under consideration the one that would most simultaneously satisfy a number of requirements for a bridge. The main requirements are: continuous and safe operation; great durability and lowest operating costs; the smallest construction cost, labor intensity of the structure, construction period, consumption of basic materials. In addition, the recommended option must comply with modern requirements and achievements in the field of industrialization of construction and comprehensive mechanization of production processes.

Reinforced concrete bridge design

For medium-sized reinforced concrete beam-split bridges across non-navigable rivers, in practice, a scheme with the same spans is often adopted. The span length in this case is one of the indicators of variation (along with the types of spans, supports, foundations).

The span length should be assigned in accordance with typical spans. In addition, it should be borne in mind that the cost of the bridge option largely depends on the span length. With high embankments, great depths of low water, weak soils on the route of the bridge crossing, due to the high cost of bridge supports, it is advisable to reduce their number by increasing the length of the spans and, conversely, with cheap supports, it is beneficial to reduce the length of the spans in order to reduce the cost of the span structures.

It should be borne in mind that according to the condition of ice-free passage of ice, the length of the spans of the channel part should be taken approximately at least 10 ÷ 15 m with a weak ice drift (ice thickness h l ≤0.5 m), 15 ÷ 20 m with average ice drift (0.5≤ h l ≤1.0 m) and 20 ÷ 30 m with strong ice drift ( h l ≥1 m).

The design of the intermediate supports can be very diverse. At the same time, it must be remembered that the use of typical supports, especially prefabricated lightweight ones, is limited by local conditions. For example, pile, post, columnar and frame intermediate supports can only be used outside the river bed and in the absence or weak ice drift. Therefore, massive supports should be used in river beds. V term paper it is recommended to use sprinkled abutments when designing. they are protected from the impact of watercourses and ice by the cone of the embankment, which in turn allows more widespread use of prefabricated lightweight structures.

Intermediate support diagram

Drawing up a diagram begins with the placement of the axes of the vertical projections of the support on which the levels of the rail base (PR), level high waters(Air-blast), low-water level (MSW), soil surface after erosion and surface of soil layers. For a given superstructure according to Appendix 1, the dimensions of the lower cushion of the support part are selected along a och and across b och bridge.

The smallest size of a reinforced concrete sub-truss slab (head) along the bridge.

l p - full length of the superstructure, m

l - design span, m

∆ - the gap between the ends of the superstructures (for reinforced concrete superstructures, 0.05 m is taken)

C 2 - the distance from the sub-farm site to the edge of the sub-farm plate, equal to 0.15 m.

Smallest sub-truss plate across the bridge

where in - distance between the axes of the beams equal to 1.8 m

b och - the size across the bridge of the lower cushion of the supporting part, m

C 1 - the distance from the bottom cushion of the support part to the edge of the sub-truss plate, taken as 0.15 ÷ 0.20 m

C 3 - the distance from the sub-farm area to the edge of the sub-farm plate, equal to 0.3 m.

The thickness of the sub-truss plate is taken as 0.8 ÷ 1.2 m.

In order to eliminate water drips on the surface of the support body, the dimensions of the support part from the bottom of the sub-truss plate to the mark corresponding to the level of high ice drift (HVL) plus 0.5 m are taken at least 0.2 m less than the dimensions of the sub-truss plate.

The underlying ice-cutting part of the support to the mark of the low ice drift level (UNL) minus the ice thickness and 0.25 m, and on the surface not covered with low-water water, 0.25 m below the soil surface after erosion, should have vertical edges and sharp edges in the plan with upstream and downstream side. Depending on the intensity of the ice drift, the angle of sharpening of the ice-cutting edge is taken in the range of 90 ÷ 120 degrees. This part of the support is taken as massive concrete. The dimensions of the ice-cutting part of the support can be taken constructively in such a way that the distance from the edge of the overlying part to the edge of the ice-breaker is at least 0.25 m.

In the course work, it is conventionally assumed that the level of low ice drift (UNL) is equal to the level of low water (LWW), and the level of high ice drift (HLW) is equal to the level of high water (HCW). The low-water level in the course work can be conventionally taken 1.5 ÷ 2.5 m below the high water level.

The pile heads are embedded in a rectangular reinforced concrete grillage 1.5 ÷ 2.0 m thick. The grillage dimensions must exceed the dimensions of the lower part of the support by at least 0.6 m. The grillage dimensions are finally determined after placing the required number of piles in it.

Air-blast = 14m; UMV = 11.5m.

BO = PR- h co; VO = 1.9-1.58 = 18.32 m;

h o = H 1 = 1.0 m;

NPP = 18.32-1.0 = 17.32 m;

VL = 14.5 m;

H 2 = NPP-VL; H 2 = 17.32-14.5 = 2.82 m;

OF = 11.5-0.85 = 10.65 m;

VL = H 3 = 14.5-10.65 = 3.85 m;

H 4 = 2.0 m;

S cr =; S cr == 1.14

V cr = 3.22;

V pr = 6.43

V 1 = a * b * c; V 1 = 1.8 * 3.36 * 1 = 6.05

V 2 = V cr + V pr; V 2 = 3.22 + 6.43 = 9.65

V 3 = 25.41

V 4 = 3.7 * 4.0 * 2.0 = 29.6

Support V = 6.05 + 9.65 + 25.41 + 20.8 = 70.71

Determination of the number of piles in the foundation of the support

Pile foundation it is advisable to use it in the construction of bridge supports, when solid soils lie at a depth of more than 5 m. In this case, the slab connecting the piles (grillage) can be buried in the ground (low pile grillage) or located above the surface of the soil (high pile grillage) after its leveling, and on rivers - above the bottom of the watercourse. Foundations with a low grillage are erected, as a rule, in dry places, for example, on floodplains of rivers or in channels if the water depth is not more than 3 m. With a deeper water depth, it is advisable to use a high pile grillage.

For intermediate supports in given soil conditions, foundations with high grillages on suspended driven reinforced concrete piles square section with dimensions 35x35, 40x40 cm.In addition, you can consider the use of hollow round piles with a diameter of 40, 50 cm with a wall thickness of 8 cm or a diameter of 60, 80 cm and a wall thickness of 10 cm. less than 5 ÷ 6 m. The length of the piles is taken as a multiple of 1 m.

The vertical loads on the pile grillage consist of the dead weight of the support parts, the pressure from the weight of the superstructure and the bridge deck, and the weight of the temporary vertical load from the rolling stock.

To determine the weight of the support itself, it is divided into parts of a simple geometric shape: a sub-truss slab, a support body above the air-blast, an ice-cutting part, a grillage. Support weight load:

where  i - normative specific gravity element material. For concrete b = 23.5 kN / m 3 for reinforced concrete reinforced concrete - 24.5 kN / m 3

V i - the volume of the support parts.

Standard load on the support from the weight of two identical spans

N ps = 24.5 * 18.9 + 4.9 * 9.3 = 508.62

where p - 4.9 kN / m - the weight of one running meter of two sidewalks with consoles and railings.

V reinforced concrete - the volume of one superstructure is taken according to Appendix 1.

Standard pressure on the support from the weight of the bridge deck

N mp = 19.4 * 2 * 9.3 = 30.70

 bp - 19.4 kN / m 3 - specific gravity of ballast with parts of the superstructure

A bp - 2 m 2 - cross-sectional area of ​​the ballast prism with track parts.

Standard pressure on the support from a temporary moving load located on two spans

with - the distance between the axes of support of adjacent superstructures.

The quantity c (Fig. 5) depends on the gap between the superstructures, as well as the total and estimated length of the superstructure and is determined in the case of using the same superstructures according to the formula:

C = 0.05 + 0.6 = 0.65

where ∆ - the gap between the ends of the superstructures

2 d - the difference between the total and calculated length of the superstructure

Total estimated vertical load on the pile grillage

N = 1 ,1(1707+508,62)+1,3*30,70+1,24*1807,84=4718,82

where γ to = 1.1 - the coefficient of reliability for the load from the weight of the structure

γ bp = 1.3 - safety factor for load from ballast weight

γ pn = (1.3- 0.003 λ) - coefficient of reliability for temporary load

The required number of piles in the support is determined by the formula:

where k g = 1.2 ÷ 1.4 - coefficient of taking into account the influence of horizontal loads

k n = 1.6 ÷ 1.65 - safety factor.

F - design bearing capacity of one pile. It is accepted depending on the type of piles according to table 4.

The resulting number of piles is placed in the plan along the grillage in an ordinary or checkerboard pattern evenly with equal distances between them in two mutually perpendicular directions. In this case, the minimum distance between the axes of the piles must be ensured, which is 3 d (d - diameter or size of the pile face). In addition, it is necessary to ensure a minimum distance from the edge of the pile to the edge of the grillage of at least 0.25 m.

If, according to these conditions, it is not possible to distribute the received number of piles in the grillage, then it is necessary to increase its size. In the event that a change in the dimensions of the grillage in the plan leads to a change in its volume, it is necessary to perform the calculation to determine the total calculated vertical load again, taking the specified dimensions of the grillage and, accordingly, clarify the number of piles.

After determining the number of bridge spans and drawing up a diagram of the bridge crossing, it is necessary to clarify the length of the piles in the intermediate supports and their number. In the case of using intermediate supports of different heights, it is necessary to perform a calculation to determine the number of piles for each of the supports. On graph paper, it is necessary to draw a diagram of the intermediate support on a scale of 1: 100.

where L about - the given hole of the bridge, m

h with - construction height of the superstructure on the support, m

l p - full length of a given superstructure, m

b - width of the ice-cutting part of the intermediate support along the bridge, m

The height of the rail foot is determined by the formula:

PR = 11.5 + 8.4 = 19.9

where is UMV - low-water level

N - the specified elevation of the rail foot above the low-water level.

The value obtained by the formula n round to the nearest higher whole number. If the fractional part of the number of spans is not more than 0.05 of the integer, then rounding is performed to the nearest smaller number of spans.

After the final designation of the bridge scheme, the distance between the cabinet walls of the abutments is calculated

L = 0.05 (6 + 1) + 6 * 9.3 = 56.15

The position of the middle of the bridge on the transition profile is determined from the condition of proportionality of the parts of the bridge opening located within the left and right floodplains.

From this condition, the distance from the middle of the river according to the low-water level to the middle of the bridge is

The sum of the widths of the ice-cutting parts of all intermediate supports

IN M - width of the river in terms of low-water level

V L, V P - width of the left and right floodplain, respectively.

The transition profile has a positive value a is deposited from the middle of the river to UMV to the right, and a negative value to the left. From the middle of the bridge in both directions is set aside 0.5 L , then the distance between the cabinet walls of the abutments is divided into spans l p + 0.05 and draw the axes of the intermediate supports.

Bridge scheme

Intermediate supports in the channel at UMV you can take the same height. On floodplains, the edge of the foundation should be located 0.25 m below the surface of the soil after erosion. The bottom of the grillage in large and medium sandy soils can be located at any level, and in heaving soils, i.e. silty, sandy loam and clayey at least 0.25 m below the freezing depth.

Depending on the height of the approach embankments and the size of the bridge spans, abutments are adopted according to standard designs (Appendix 2). The slope of the embankment cone with a slope of 1: 1.5 should pass below the abutment under-farm platform by at least 0.6 m. The embankment edge is placed 0.9 m below the rail foot.

The following dimensions must be indicated on the facade of the bridge:

• the length of the bridge (the distance between the back faces of the abutments);
• the length of the spans and the size of the gap between the ends;
• the elevation of the bottom of the structure (NK), which must be at least 0.75 m higher than the air-blast;
• the mark of the levels of high and low-water waters, the base of the rail (PR), the edge of the embankment (BN), the top of the support (VO), the cut-off (OF) and the base of the foundation (PF);

Bibliography

1. SNiP 2.05.03-84. Bridges and pipes / Gosstroy of the USSR. Moscow: TsITP Gosstroy USSR, 1985 .-- 253 p.
2. Manual to SNiP 2.05.03-84 "Bridges and pipes" for the survey and design of railway and road bridge crossings over watercourses (PMP-91) Moscow 1992
3. SNiP 3.06.04-91 Bridges and pipes / Gosstroy of the USSR. M .: TsITP Gosstroy USSR, 1992 .-- 66 p.
4. GOST 19804-91 Reinforced concrete piles. Technical conditions.M .: TsITP Gosstroy USSR, 1991 .-- 15 p..
5. Kopylenko V.A., Pereselenkova I.G. Design of a bridge crossing at the intersection of a river by a railway line: Textbook for higher educational institutions of the railway. transport / Ed. V.A. Kopylenko. - M .: Route, 2004 .-- 196 p.
6. Design of bridge crossings on railways: Textbook for universities / M.I. Voronin, I.I. Kantor, V.A. Kopylenko and others; Ed. I.I. Cantor. - M .: Transport, 1990 .-- 287 p.
7. Bridges and tunnels on railways: Textbook for universities / V.O. Osipov, V.G. Khrapov, B.V. Bobrikov and others; Ed. IN. Osipova. - M .: Transport, 1988 .-- 367 p.

The load-bearing elements of the carriageway - reinforced concrete slabs of the carriageway (assumed to be 18 cm thick) take the load from vehicles from the roadbed, from pedestrians from sidewalks and transfer them to the main load-bearing structures of the span.

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Ministry of Education of the Russian Federation

Federal State Budgetary educational institution higher professional education

"Siberian State Automobile and Highway Academy (SibADI)"

Department "BRIDGES"

Course project

"Design of a road reinforced concrete bridge»

Completed:

Student ADb-12- Z 1 group

Zhdanov A.V.

Accepted:

Shchetinina N.N.

Omsk - 2014

1. Description of the bridge layout and superstructure design _____________ 2

2. Calculation of the roadway slab _______________________________________ 4

2.1. Determination of forces in the slab of the carriageway from a constant load ___4

2.2. Determination of efforts from live load ________________________5

2.3. Reinforcement of the PCh slab and strength calculation _____________________10

2.3.1. Reinforcement of the PCh slab in the middle of the slab _______________________11

2.3.2. Reinforcement of the PCh slab on supports ______________________________12

3. Calculation and design of the main beam ____________________________ 14

3.1. Determination of forces in a beam from a constant load ______________14

3.2.1. Accounting for spatial work ________________________________15

3.2.2. Definition of the KPU ___________________________________________16

3.3. Determination of forces in the main beam ___________________________18

3.4. Reinforcement of the main beam ___________________________________25

4. Plotting materials ____________________________________27

5. Calculation of the inclined section for the shearing force _________________28

List of used literature __________________________________ 30

Appendix 1_______________________________________________31

Appendix 2_______________________________________________32

1. Description of the scheme of the bridge and the design of the superstructures.

Bridge crossing– this is a complex of structures, which includes a bridge, approaches to it; as well as ice cutters, regulation structures and bank protection devices, which are not represented in the project.

The bridge with its structures covers the channel and part of the river floodplain. The bridge consists ofspan structures and supports.

SpansThe bridge includes the following main parts: carriageway, bearing part (beams), connection system and support parts.

Carriageway perceives the action of moving loads (from vehicles and pedestrians) and transfers them to the bearing part. The roadway includes a bridge deck and load-bearing elements.

In accordance with the task, the overall dimension of the bridge is G10 (for technical category III), the roadbed consists of two lanes: the width of the carriageway is 7.0 m, and the width of the road is 2x1.5 m. The width of the bridge, including the width the carriageway, safety lanes, sidewalks and fences is equal to:

The width of the sidewalk, according to the assignment, is 2.25 m.On the outside, the sidewalks are fenced with handrails with a height of 1.1 m, and on the inside with a barrier fence with a height of 0.75 m.To ensure quick drainage of water, we attach a longitudinal slope to the surfaces of the roadbed and sidewalks (10 ‰) and cross slopes (20 ‰). The need to ensure a smooth transition from the embankment to the bridge is achieved by creating special transition sections in the form of transition plates at the interface between the bridge and the embankment.

The load-bearing elements of the carriageway - reinforced concrete slabs of the carriageway (assumed to be 18 cm thick) take the load from vehicles from the roadbed, from pedestrians from sidewalks and transfer them to the main load-bearing structures of the span. The bearing part of the superstructure perceives the action of the superstructure's own weight and the temporary moving load and transfers it to the supports, which are beams.

The bridge bed ensures the safe movement of vehicles and fencing devices, devices for drainage systems, expansion joints and interfaces of bridges with approaches.

1 - asphalt concrete pavement - 9 cm;

2 - protective layer - 6 cm;

3 - waterproofing - 0.5 cm;

4 - leveling layer - 3 cm;

5 - reinforced concrete slab-18 cm

Figure 1.3. - Cross section of the main beam.

2. Calculation of the roadway slab

1. Determination of forces in the slab of the carriageway

from constant load.

Determination of the design load acting on 1 m 2 carriageway slabs (dead weight) are presented in Table 1.1.

; ; (SNiP table 8)

Determination of the design load

Table 1.1.

Calculated maximum bending moment in the middle of the span of the slab M q and the calculated maximum shear force Q g on a support from a constant load are equal:

M q = q p * l p 2;

Q q = q p * l p;

where

l p - calculated span of the slab, l p = l - b p;

1 - the distance between the axes of the beams;

b p - beam rib width.

2.2. Determination of efforts from live load

I determine the calculated distance between the beams:

Where l o - the distance between the axes of the beams;

b p - rib thickness.

Determination of efforts from the load A-11.

Figure 2.1 - Diagram of working widths for determining the maximum bending moment when loaded with A14 load.

Since the calculated distance between the beams is less 2m , then when determining the efforts from the live load A-14, consider the layout of one track and one load wheel (Fig. 2.1).

Wheel pressure on the surface of the pavement, acting on the site a b , spread by pavement at an angle of approximately 45 °. As a result, pressure is transferred to the surface of the reinforced concrete slab over a much larger area (working width diagram). In shape, it is mistaken for rectangular.

When determining the bending moment, the load is placed symmetrically relative to the roadway slab.

We accept the common area of ​​pressure distribution:

a 1 = a + 2 h to = 0.2 + 2 0.185 = 0.57 m

b 1 = b + 2 h to = 0.6 + 2 0.185 = 0.97 m

where H = 0.185 m - the thickness of the pavement layers

2 from the trolley and from the distributed strip:

Determine the reliability factors for the load:

 fa T  fa T = 1.5;

 fa  fa = 1.15.

- dynamic coefficient;

Determine the maximum bending moment in the middle of the span of the roadway slab:

Total moment from constant and temporary loads:

Figure 2.2 - Diagram of working widths for determining the maximum lateral force when loaded with load A14.

When determining the lateral force, the load is set so that the edge of the pressure distribution area coincides with the tested section (Fig.2.2)

The dimensions of the diagrams of working widths have the same meaning as when determining the magnitude of the bending moment. The load safety factors remain the same.

Maximum lateral force at support:

where y 1 = 0.74 Is the ordinate of the line of influence under the wheel axis.

Total shear force from constant and temporary loads

Determination of efforts from the load NK-100

Figure 2.3 - Diagram of working widths for determining the maximum bending moment when loaded with a load of NK-100.

Under the action of a load from one wheel, the dimensions of the platform will be:

along the movement a 3 = a 1 = 0.57 m;

across traffic b 3 = b + 2H = 0.8 + 2 · 0.185 = 1.17 m.

When determining the bending moment, the load is placed in the middle of the span (Figure 2.3)

I determine the dimensions of the plot of working widths, choosing the largest of the two values:

Determine the intensity of the distributed load per 1m 2 : .

- dynamic coefficient,;

- coefficient of reliability for the load.

Determine the maximum bending moment in the middle of the span:

Total bending moment from constant and temporary loads:

Figure 2.4 - Diagram of working widths for determining the maximum lateral force when loaded with a load of NK-100.

When determining the lateral force, the load is placed as close as possible to the edge of the beam (Figure 2.4)

Determine the magnitude of the lateral force:

where y 1 = 0.69 - ordinate of the line of influence along the wheel axis.

Total shear force from constant and temporary loads:

As the design efforts, the largest ones are taken, obtained when the slab is loaded with a load of A-14:

We determine the moments for the actual loading scheme:

M 0.5 l = 0.5 M max = 0.5 43.21 = 21.61 kN m;

M op = -0.8 M max = -0.8 43.21 = -34.57 kN m.

3. Calculation and design of the roadway slab.

Based on the calculated values ​​of the efforts, we make the reinforcement of the roadway slab and check it for strength.

1. Bottom mesh reinforcement

The scheme for calculating the lower grid is shown in Figure 2.5.

Rice. 2.5 - Scheme for calculating the lower grid

1. z ≈ 0.925 h o = 0.925 0.155 = 0.1434 m.

PCS. I accept 6 rods.

M pre = 18.6 kNm> M 0.5 l = 17.73 kNm.

Therefore, the strength test condition is satisfied.

Determine the number of distribution reinforcement bars:

PCS. We accept 4 rods constructively.

Actual area of ​​distribution valves, A s f ':

M 2.

2.3.2. Reinforcement of the PCh slab on supports (upper mesh).

The scheme for calculating the upper grid is shown in Fig. 2.6.

1. Determine the working height of the slab:

1. Determine the shoulder of the internal pair of forces:
   z ≈ 0.925 h o = 0.1156 m.

1. Determine the area of ​​the working reinforcement:

4. Determine the number of rods:

PCS. We accept 12 rods constructively.

I determine the actual area of ​​the working reinforcement:

1. Determine the height of the compressed zone:

1. I carry out a strength check:

M pre = 29.2 kNm> M op = 28.36 kNm, hence the strength test condition is met.

1. Determine the area of ​​distribution fittings:

We accept the diameter of the distribution fittings: d '= 6 mm

2. Determine the number of rods of distribution reinforcement:

PCS. We accept 7 rods.

3. Actual area of ​​distribution fittings, A s f ':

M 2.

3. Calculation and design of the main beam.

3.1. Determination of forces in a beam from a constant load

The permanent load is determined per 1 running meter. beams and is composed of the weight of the beam itself, the roadway slab, pavement, litas, curb stones and railings.

The determination of forces from a constant load is made in tabular form and is shown in Table 3.1.

Table 2.1. Calculation of the permanent load on the main beam

Amount 189.05

We believe that the constant load is distributed evenly between all beams and the load on each of them is equal to:

kN / m 2.

1. Determination of transverse installation factors

The distribution of the temporary vertical load between the main beams is carried out using the transverse installation factor (KPU), which shows how much of the temporary load on the roadway and sidewalk falls on the calculated beam.

KPU is determined by the eccentric compression method. To determine the transverse installation, it is necessary to build lines of influence of the forces acting on individual beams.

In view of the straightness of the pressure influence lines, to construct them, it is enough to find two ordinates above the extreme beams:

Or.

thus: y 1 = 0.42, y 8 = -0.17.

To determine the forces in the main beam from the live load, it is necessary to find the KPU along the line of the influence of pressure on the calculated beam. At the same time, for the load A-11 for the bogie and the strip, the KPU is determined differently. In this case, a combination factor is introduced for the strip, equal to 0.6 for the second column.

For trolley

For evenly distributed strip

From the crowd

The section is loaded where we have a positive value of the effort.

3.2.2. Determination of the KPU for the main beam

1st loading scheme.

Load A11 is placed 1.5 m from the safety lane with one loaded sidewalk.

Rice. 3.1 - Scheme of loading the pressure influence line with a load A11 according to I loading scheme

2nd loading scheme.

Load A11 is placed 0.55 m from the curb with unloaded sidewalks.

Rice. 3.2 - Scheme of loading the pressure influence line with a load A11 according to II loading scheme

I determine the coefficients of the transverse installation:

3rd loading scheme.

One NK-80 calculated vehicle is placed as close as possible to the safety lane with unloaded sidewalks.

Rice. 3.3 - Scheme of loading the pressure influence line with the NK-80 load.

I determine the coefficient of transverse installation:

3.3. Determination of forces in the main beam

Calculated force values M and Q are determined by loading the influence lines with constant and temporary loads. Determine the values ​​of M and Q in sections, the number of which is sufficient to construct diagrams of these efforts: the middle, quarter and support section of the beam.

Force in the considered section:

Where

S - effort in the section under consideration;

q p –Designed permanent load per 1 running meter. main beam = 23.63 kN / m 2 ;

 - the algebraic sum of the areas of all loading areas of the influence line;

 - the area of ​​the line of influence with a positive value;

 fv - safety factor for the strip; fv = 1,2

 v - coefficient of lateral installation for a strip of automobile load;

- dynamic coefficient for loads А11 and НК-80;

 P - safety factor for the trolley;

 P = 1.5 for  = 0,  p = 1.2 for  ≥ 30 m, intermediate values ​​- by interpolation:

γ f NK-80 - safety factor for NK-80 load= 1;

 P - coefficient of transverse installation for the bogie;

 NK-80 - coefficient of transverse installation for the load trolley NK-80;

P axis - efforts on the bogie axle A11 = 108 kN;

r NK-80 - efforts on the load axis NK-80 = 20 t;

y 1, y 2, y 3, y 4 - the ordinates of the line of influence for the axes of the load;

 T - safety factor for pedestrians; f Т = 1,2

 T - coefficient of transverse installation for pedestrians;

l p = 8.4 m - calculated span length.

Rice. 3.4 - Scheme of loading lines of influence of forces M and Q I loading scheme.

Rice. 3.5 - Scheme of loading lines of influence of forces M and Q constant and temporary loads in sections 1-1,2-2 and 3-3 along II loading scheme.

Rice. 3.6 - Scheme of loading lines of influence of forces M and Q permanent and temporary NK-80 loads in sections 1-1,2-2 and 3-3.

Section 1-1

I define M

1 i loading scheme

2 i loading scheme

3rd i loading scheme

Determine Q

1 i loading scheme

2 i loading scheme

3rd i loading scheme

Section 2-2

I define M

1 i loading scheme

2 i loading scheme

3rd i loading scheme

Determine Q

1 i loading scheme

2 i loading scheme

3rd i loading scheme

Section 3-3

The moment in the reference section is zero.

Determine Q

1 i loading scheme

2 i loading scheme

3rd i loading scheme

The calculation results are summarized in Table 3.2.

Table 3.2.-Internal forces on sections

On the basis of the calculation made, I determine the maximum forces in the sections and the structure of the diagram of the enveloping forces (Fig. 3.7).

Rice. 3.7. - Diagram of enveloping efforts

1. Reinforcement of the main beam.

Rice. 3.8 - Designation of the calculated width of the slab.

A s (A ’s ) - area of ​​stretched (compressed) reinforcement;

a s (a ’s ) - distance to central t. stretched (compressed) reinforcement;

h = 0.9 m - the height of the design beam;

h f = 0.18 m - the height of the slab of the carriageway of the beam;

b = 0.16 m - beam rib thickness;

1. Calculated slab width

1. Inner pair shoulder:

1. Working reinforcement area:

m 2;

1. Number of rods per rod diameter d = 22 mm:

PCS., round up n s f = 8 pcs.

Actual area of ​​working reinforcement:

m 2.

5. Position of the center of gravity:

where n s - the total number of rods; n i - the number of rods in i -th row; a i - distance to center

severity i -th row from the bottom of the beam;

6. Exact calculation of the working height:

7. Height of the compressed zone:

(m);

Working condition factor:

where: (h - x ) Is the height of the stretched sectional zone; - distance from the axis of the stretched reinforcing element from the stretched face of the section;

We accept

Limiting torque check:

M pr> M max; 653.03> 551.08

Therefore, the reinforcement is calculated correctly.

Figure 3.9- Scheme for checking the strength of the beam at the limit moment.

4. Plotting materials.

1. A diagram of moments is constructed ( M max ), postponing the limiting moment М pre> M max within 5%
2. The limiting moment is divided by the number of pairs of rods.

1. According to SNiP (p 3.126), we determine the value of the rod termination:

For concrete grade B30 l s = 22 d = 22 0.022 = 0,

484m

1. The rods are bent at an angle of 45 °. The bent bars should be distributed along the length of the beam in such a way that any section normal to the axis of the element intersects at least one bar; if this requirement is not met, then we use additional oblique rods, welded to the main working reinforcement (of the same diameter).

The length of the welded seams at the points of attachment of the inclined rods is taken to be equal for one-sided welding - 12d, for double-sided - 6d.

In places where the bending or breaking of the rods is made, as well as between them at distances not exceeding ¾ of the beam height, it is necessary to place tie seams in the welded frames. Their length is assumed to be 6d and 3d. In double-sided welding, the smallest thickness of the seams is 4 mm (p. 3.161).

5. Calculation of an inclined section for shearing force.

Figure 5.1 - a diagram for calculating the strength of a beam along an inclined section

We carry out the calculation of the support section:

1. The calculation of an inclined section of a member with transverse reinforcement for the action of a shear force should be made from the condition:

where: - cross-sectional area of ​​one bend bar; - coefficient of working conditions; - the number of bends caught in the inclined section; - the number of slices; - the angle of inclination of the bent rods to the longitudinal axis of the element at the intersection of the inclined section;

MPa

where: - cross-sectional area of ​​one rod of the clamp; - coefficient of working conditions; - the number of clamps caught in the inclined section; - the number of slices;

6 clamps;

MPa

but not less than 1.3 and not more than 2.5;

design resistance shear when bending; the greatest shear stress from the standard load;

Pa

kN;

kN;

The check condition is met.

where: area of ​​horizontal non-tensioned reinforcement, cm 2 ;

Since hail, then K<0 и он не учитывается.

6.MPa - the check is in progress.

The calculation is correct.

List of used literature:

1. Kolokolov N.M., Kopats L.N., Fainshtein I.S. Artificial constructions:

Textbook for technical schools transp. p-va / Ed. N.M. Kolokolov. - 3rd ed.,

Revised and additional - M .: Transport, 1988, 440s.

2. Bridges and structures on the roads: Textbook. for universities: In 2 hours / Gibshman E.E.,

Kirilov V.S., Makovsky L.V., Nazarenko B.P. Ed. 2nd, rev. and add. –M .:

Transport, 1972, 404s.

3. Bridges and structures on the roads: Textbook. for universities: In 2 hours / P.M. Salamakhin,

O.V. Volia, N.P. Lukin and others; Ed. P.M. Salamakhin. -M .: Transport, 1991,

344s.

4. Design of wooden and reinforced concrete bridges. Ed. A.A.

Petropavlovsky. Textbook. for universities .- M .: Transport, 1978, 360s.

5. SNiP 2.05.03-84 *. Bridges and pipes - Moscow: Stroyizdat, 1984

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1. Design assignment

2. Development of options

2.1 Option 1

2.1.1 Defining the bridge layout

2.2 Option 2

2.2.1 Defining the bridge layout

2.2.2 Determination of consumption of building materials and cost

2.3 Comparison of options

3. Calculation of the superstructure

3.1.2 Calculation of slab sections

Literature

1. Design assignment

Develop a project of a reinforced concrete bridge for a single track railroad 42 m hole across the river. Construction area - Leningrad region. The elevation of the edge of the embankment - 17.90 m.Mezhenny water level (LWL) - 10.10 m, high water level (HCL) - 13.00 m.Low ice drift level (UNL) - 10.30 m, high ice drift level - 12 , 80 m, ice thickness - 0.4 m. Standard temporary vertical load - C13. The total erosion coefficient is K = 1.3.

The transition profile with indication of water horizons and engineering geology is shown in the diagram (Fig. 1.1).

2. Development of options

2.1 Option 1

2.1.1 Defining the bridge layout

Assuming the use of crumbling abutments and taking into account that the opening of the bridge is 42 m, a two-span bridge layout with sectional standard beams 227.6 is planned. The required length of the bridge between the extreme points of the abutments:

Ln = l0 + nb + 3H + 2a, where

n is the number of bulls falling into the water;

b is the thickness of the intermediate support at the level of the high water horizon, m;

a - the value of the entry of the abutment structure into the embankment, m;

H is the height of the embankment from the middle line of the trapezoid formed by the horizons of high and low-water waters (along which the bridge opening is measured) to the edge of the embankment, m;

3H - the length of two slopes of the embankment cones with a steepness of 1: 1.5;

l0 - bridge hole.

Taking into account the presence of ice drift, streamlined bulls, prefabricated-monolithic, 2.6 m thick, were taken in relation to typical project 3.501-79.

The number of supports falling into the water is equal to one (n = 1). Under these conditions, the required length of the bridge on top will be:

Ln = 42 + 2.6 + 3? 6.35 + 2? 0.75 = 65.15 m.

Pile abutments are accepted. The length of the abutment wing on top when the adjacent beams span 27.6 m is 5.3 m. Taking into account the distance between the ends of the beams by 0.05 m, the actual length of the bridge with the adopted structures will be:

Lf = 2? 27.6 + 2? 5.3 + 3? 0.05 = 65.95 m.

This length is less than the required

2.2.2 Determination of consumption of building materials and cost

Spans.

The volume of reinforced concrete of the superstructure with a total length of 27.6 m with a ride on top of 83.0 m3.

Intermediate support.

a support with a height of 5.3 is taken in the form of prefabricated monolithic structures.

Determine the required number of hollow reinforced concrete pillars with a diameter of 100 cm and a length of 15 m, filled after immersion concrete mix... The number of piles is calculated using the formula:

m - coefficient taking into account the influence of the bending moment acting on the base of the grillage, equal to 1.5 - 1.8, and the greater the influence of constant centrally applied forces in the value of UN, the lower the value of the coefficient m;

УN - the sum of the calculated vertical forces acting on the base of the foundation, tf.

УN = УNvr + УNball + УNpr.p. + УNop, where

UNvr, UNball, UNpr.str., UNop - vertical pressures, tf, respectively, from the temporary load when loading two adjacent spans, from the weight of the ballast on the spans of the railway bridge, from the weight of reinforced concrete spans and from the weight of the support with the foundation.

The indicated values ​​are determined by the formulas:

N. p. = 1.1? Vpr. Page? 2.5? 0.5; Nop = 1.1? Vop? 2.4;

l1 and l2 are the total lengths of the spans resting on the supports, m;

d - safety factor for temporary load;

2.0 - volumetric mass of ballast, t / m3;

1.3 - safety factor for ballast;

Fball - ballast trough cross-sectional area, m2;

1.1 is the safety factor for the dead weight of the structure;

Vpr. P. - the volume of reinforced concrete of the superstructures resting on the support, m3;

2.5 - bulk density of reinforced concrete, t / m3;

Vop - the volume of the support body and foundation, m3;

Pd - design bearing capacity of one pile (shell-pile).

N. p. = 1.1? (83 + 83).? 2.5? 0.5 = 228.3 tf;

Nop = 1.1? (30.3 + 46.5 + 48.8)? 2.4 = 331.6 tf;

YN = 405 + 129.2 + 228.3 + 331.6 = 1094.1 tf.

The volume of hollow pillars with a wall thickness of 10 cm at the rate of 8 pieces per support

The volume of reinforced concrete of the abutment head is 46 m3.

Volume of concrete for filling hollow piles

The scope of work and the determination of the cost of structural elements are shown in Table 2.1.

Table 2.1 Determination of the cost of structural elements for option 1

Determination of the total cost of the bridge is given in table 2.2.

Table 2.2 Determination of the total cost for option 1

2.2 Option 2

2.2.1 Defining the bridge layout

The number of supports falling into the water is equal to two (n = 2). Under these conditions, the required length of the bridge on top will be:

Ln = 42 + 2 2.6 + 3? (17.9 -0.5 (13.0 - 10.1) +10.1) + 2? 0.75 = 67.8 m.

Pile abutments are accepted. The length of the abutment wing on top with the span of the adjoining beams 16.5 and 27.6 m is 5.3 m. Taking into account the distance between the ends of the beams by 0.05 m, the actual length of the bridge with the adopted structures will be:

Lf = 16.5? 2 + 27.6 + 5.3 +3.75 + 4? 0.05 = 69.85 m.

This length is longer than required by

2.2.2 Determination of consumption of building materials and cost

Spans.

The volume of reinforced concrete of the superstructure with a total length of 16.5 m with a ride on top of 35.21 m3. The volume of reinforced concrete of the superstructure with a total length of 27.6 m with a ride on top of 83.00 m3

Intermediate supports

2 supports with a height of 5.3 and 6.7 m are taken in the form of prefabricated monolithic structures.

The volume of reinforced concrete support blocks 5.3 m high is approximately

The concrete for grouting the blocks and the concrete for filling the supports with a height of 5.3 m is approximately

The volume of reinforced concrete support blocks with a height of 6.7 m is 38.3 m3

The concrete for grouting the blocks and the concrete for filling the supports with a height of 6.0 m is

2.4 + 56.4 = 58.8 m3

The volume of the grillage with a height of 1.6 m from monolithic reinforced concrete will be taken with dimensions in terms of 8.63.6 with bevels (to improve flow conditions) of 0.5 m each:

1.6? (3.6? 8.6 - 4? 0.5? 0.5? 0.5) = 48.8 m3.

The required number of hollow pillars of centrifuged reinforced concrete with a diameter of 100 cm and a length of 13 m, filled after immersion with a concrete mixture, is determined.

For an intermediate support with a height of 5.3 m, the following was obtained:

N. p. = 1.1? (35.21 + 83.00).? 2.5? 0.5 = 162.5 tf;

Nop = 1.1? (30.3 + 46.5 + 48.8)? 2.4 = 331.6 tf;

YN = 360.6 + 103.2 + 162.5 + 331.6 = 957.9 tf.

The bearing capacity of each pillar with a diameter of 100 cm and a length of 13 m of the pile on the ground Rd is about 220 tf.

Let's take 8 pillars with a diameter of 100 cm and a length of 13 m under a support.

Volume of concrete for filling hollow piles

For an intermediate support with a height of 6.7 m, the following was obtained:

N N ex p. = 1.1? (35.21 + 35.21).? 2.5? 0.5 = 290.5 tf;

Nop = 1.1? (58.8 + 38.3 + 48.8)? 2.4 = 385.2 tf;

YN = 293 + 77.2 + 290.5 + 385.2 = 957.9 tf.

The bearing capacity of each pillar with a diameter of 100 cm and a length of 15 m of the pile on the ground Рд is about 250 tf.

Let's take 8 pillars with a diameter of 100 cm and a length of 15 m under a support.

The volume of hollow pillars with a wall thickness of 10 cm at the rate of 8 pieces per support

Volume of concrete for filling hollow piles

The volume of reinforced concrete of the head of the abutment for the superstructure 16.5 m long is 40.0 m3, the abutment for the superstructure with a length of 27.6 m is 46.0 m3.

The volume of 9 hollow piles with a diameter of 60 cm and a length of 10 m with a wall thickness of 10 cm.

Volume of concrete for filling hollow piles

The scope of work and the determination of the cost of structural elements are shown in Table 2.3.

Table 2.3 Determination of the cost of structural elements for option 2

Determination of the total cost of the bridge is given in table 2.4.

Table 2.4 Determination of the total cost for option 2

2.3 Comparison of options

According to option 1, capital costs will amount to 119.4 thousand rubles, according to option 2 - 148.0 thousand rubles.

When switching from a two-span scheme to a three-span one, the total cost increased by 28.6 thousand rubles. Option 1 is accepted for calculation.

3. Calculation of the superstructure

The calculation is carried out for a typical girder two-block superstructure with a length of 23.6 m made of prestressed concrete with ballast riding (Figure 3.1, a).

3.1 Calculation of the carriageway of the superstructure

3.1.1 Determination of design forces

Rice. 3.1 Design scheme of the roadway slab

The outer and inner slab works under a vertical load as a cantilever clamped by one side in the rib of the beam (Fig. 3.1). On the inner console, the loads are considered uniformly distributed along the entire length, and on the outer console, the distribution of loads on sections of different lengths and the action of concentrated forces from the mass of the railings and sidewalks are taken into account.

lk = 0.9 - 0.13 = 0.77 m;

l1 = 1.7 - 0.9 - 0.13 = 0.67 m; l2 = 1.99 - 0.9 - 0.13 = 0.96 m;

l3 = 2.09 - 0.9 - 0.13 = 1.06 m; l4 = 2.66 - 0.9 - 0.13 = 1.63 m;

0.5bt = 0.285 m.

Standard constant loads at the calculated width of the slab section along the span from the dead weight:

one-sided metal railings Pp = 0.07 tf / m;

reinforced concrete slab of the sidewalk Рт = hтbтгжб = 0.1? 0.57? 2.5 = 0.14 tf / m;

ballast trough plates qpl = hplgzhb = 0.2? 2.5 = 0.5 tf / m2;

ballast with parts of the track qb = hbgb = 0.5? 2.0 = 1.0 tf / m2.

The temporary uniformly distributed load from the mass of track materials and ballast, stacked on the sidewalk during track repair, is taken as pb = 1.0 tf / m2. This load is not taken into account together with the temporary load from the rolling stock.

The overload coefficient of constant loads Pp, Pt, qpl and temporary load pb is taken n1 = 1.1, constant load qb - n2 = 1.3. The overload coefficient of the temporary load from the rolling stock and the dynamic coefficient 1 + m are calculated by the formulas:

nvr = 1.3 - 0.003l = 1.3 - 0 = 1.3;

Efforts when calculating strength:

for the outer console in section 1 under the action of a temporary load from the rolling stock

Q1 = n1 (Pp + Pt + qpll3) + n2qbl2 + nvr (1 + m) pl1 = 1.1 (0.07 + 0.14 + 0.51? 1.06) + 1.3? 1.0? 0.96 + 1.3? 1.5? 7.65? 0.67 = 12.1 tf / m.

for the outer console in section 1 under the action of a temporary load from the mass of track materials and ballast temporarily folded on the sidewalk

Q1 = n1 (Pp + Pt + qpll3) + n2qbl2 + n1pbbt = 1.1 (0.07 + 0.14 + 0.5) + 1.3? 1.0? 0.96 + 1.1? 1, 0? 0.57 = 2.67 tf / m.

Q2 = lk =? 0.77 = 12.91 tf / m.

Efforts for Fatigue and Crack Opening.

Calculated with overload factors n1 = n2 = nvr = 1.0 and dynamic factor:

for the outer console in section 1

Q "1 = Pp + Pt + qpll3 + qbl2 + (1 + m) pl1 = 0.07 + 0.14 + 0.5? 1.06 + 1.0? 0.96 + 1.33? 7.65 ? 0.67 = 8.52 tf / m.

for the inner console in section 2

Q "2 = lk =? 0.77 = 8.99 tf / m.

3.1.2 Calculation of slab sections

The calculation of the slab is carried out for strength, endurance and crack resistance. Slab sections are designed for forces M and Q, as defined in section 3.1.1.

Strength calculation.

The rectangular section of the slab has an estimated width b = 1.0 m (Fig. 3.2, a). Slab thickness hpl = 0.20 m.

We set the working reinforcement of a periodic profile of class A-III with a diameter of d = 14 mm (design resistance to strength Ra = 3100 kgf / cm2, design resistance to endurance at c = 0 Ra = 1800 kgf / cm2, cross-sectional area f = 1.54 cm2 ). The concrete grade of the M500 slab (the design compressive strength when calculating the strength Rпр = 235 kgf / cm2, the design compressive strength when calculating the endurance R "пп = 175 kgf / cm2, the conditional main tensile stresses Rg.r.d. = 42 kgf / cm2).

Useful (working) section height with a protective layer thickness of 2 cm

h0 = hpl - 0.5d - 2 cm = 20 - 0.5? 1.40 - 2.0 = 17.3 cm.

The required height of the compressed zone in the ultimate strength state (with a rectangular plot of stresses in concrete):

Rice. 3.2 Design diagrams of the cross-section of the slab: a - when calculating the strength; b - when calculating for endurance; c - when calculating for crack resistance

Required reinforcement in the tensioned zone of the slab

Number of rebars

8 reinforcement bars are accepted per linear m of the slab width. Then the area of ​​the reinforcement will be

Fa = 8f = 8? 1.54 = 12.32 cm2.

Compressed zone height

z = h0 - 0.5x2 = 17.3 - 0.5? 1.62 = 16.5

Checking the strength of the section by bending moment

Mпр = Rпрbx2 (h0 - 0.5x2) = 2350? 1? 0.0162? (0.173 - 0.5? 0.0162) = 6.28 tf? M;

M = 4.97 tf? M;

The check is in progress.

Endurance calculation.

Fatigue calculations are made on the assumption that the material of construction is elastic. The concrete of the stretched zone is not taken into account in the calculation (Fig. 3.2, b). The maximum stresses in the compressed zone of concrete and tensile reinforcement are compared with the design resistances. Design resistances of materials are set depending on the characteristics of the cycle of acting stresses

The height of the compressed zone of the reduced section

For concrete grade M500, the ratio of the moduli of elasticity of reinforcement and concrete under a multiply repeated load n "= 10, then

Shoulder of a pair of internal forces with a triangular diagram of compressive stresses in concrete

Stress testing is carried out according to the formulas

in fittings

For external console

for c = 0.31

ksb = 1.052, kca = 1.21,

ksbR "pr = 1.052? 145 = 183.75 kg / cm2, ksbR" a = 1.21? 1800 = 2178 kg / cm2,

in fittings

For interior console

for c = 0.13

ksb = 1.006, ks = 1.065,

ksbR "pr = 1.006? 145 = 176.05 kg / cm2, ksbR" a = 1.065? 1800 = 1917 kg / cm2,

in fittings

All checks are carried out.

Crack resistance calculation.

The calculation limits the opening of normal cracks and the magnitude of tensile stresses in concrete. Determination of the opening of normal cracks: the area of ​​the interaction zone of reinforcement with concrete (Figure 3.1, c)

Fr = b (a + 6d) = 100 (2 + 6? 1.4) = 1040 cm2;

reinforcement radius

stresses in the working armature

the condition must be met

D = 0.02 cm - ultimate crack opening,

Ea = 2.1 × 106 kgf / cm2 is the elastic modulus of the bar reinforcement,

w2 = 0.75 - coefficient taking into account the effect of tensile concrete on the deformation of reinforcement, for concrete grade M500,

Checking the main tensile stresses at the level of the neutral axis:

is carried out for the shear force in the design section from the standard loads, taking into account the dynamic coefficient

All checks are carried out. Because ug.r. ? 0.7Rp = 8.75 kgf / cm2, calculation for inclined sections is not performed.

3.2 Calculation of the main beams of the superstructure

3.2.1 Determination of the design forces in the main beam

The constant load on the superstructure consists of the dead weight of the structure and the weight of the bridge deck.

Standard load for 1 linear meter m of the main beam is determined: from its own weight

from the mass of the bridge deck with ballast riding

Efforts when calculating strength. The overload factors in the strength calculation are taken for the dead weight of the structure 1.1, for the weight of the bridge deck with ballast riding 1.3, for the standard load nvr = 1.3 - 0.003l = 1.3 - 0.003 26.9 = 1, 22. Dynamic coefficient

effort M1

M1 = w1 = (1.1 3.76 + 1.3 1.80 + 1.22 1.21 7.78) 67.84 = 1218.5 tf m;

effort M2

M2 = w2 = (1.1 3.76 + 1.3 1.80 + 1.22 1.21 7.26) 90.45 = 1555.1 tf m;

Q0 = w3 = (1.1 3.76 + 1.3 1.80 + 1.22 1.21 8.30) 13.45 = 251.9 tf;

Q2 = nvr (1 + m) p4sh4 = 1.22 1.21 10 3.36 = 49.6 tf.

Efforts when calculating fracture toughness.

The overload factors in the strength calculation are taken for the self-weight of the structure 1, for the mass of the bridge deck with ballast riding 1, for the standard load

Dynamic coefficient

Total forces in the sections of the calculated beam:

force M "1

M "1 = w1 = (3.76 + 1.80 + 1.23 7.78) 67.84 = 1026.4 tf m;

force M "2

M "2 = w2 = (3.76 + 1.80 + 1.23 7.26) 90.45 = 1307.3 tf m;

force Q "0

Q0 = w3 = (3.76 + 1.80 + 1.23 8.3) 13.45 = 212.1 tf;

force Q "2

Q "2 = (1 + m) p4sh4 = 1.23 10 3.36 = 41.3 tf.

3.2.2 Bending moment strength analysis

The bending moment strength calculation was carried out for the most loaded section 2-2 (M2 = 1555.1 tf m).

We set a working reinforcement made of high-strength smooth wire of class Bp-II with a diameter of d = 5 mm (standard resistance = 17000 kgf / cm2, calculated tensile strength Rn2 = 10100 kgf / cm2, cross-sectional area f = 0.196 cm2). Reinforcement is performed with wire bundles of 24 wires (cross-sectional area f = 24 0.196 = 4.704 cm2, diameter of closed channels d = 5 cm). Concrete grade of the M500 slab (design compressive strength when calculating the strength Rпр = 235 kgf / cm2, spalling during bending Rsc = 45 kgf / cm2, compression when calculating the combined effect of force factors and adverse effects of the external environment Re = 205 kgf / cm2, conditional principal tensile stresses Rg.r.p. = 42 kgf / cm2, tension when calculating the formation of cracks Rr.p. = 17.5 kgf / cm2, compression when calculating for longitudinal crack resistance Rt = 225 kgf / cm2).

The actual T-shape of the cross-section is replaced by the calculated one (Fig. 3.5).

Calculated slab width

Actual slab area with heaps:

Estimated shelf height

Approximate distance from the lower edge of the chord to the center of gravity of the reinforcement

Working section height

h0 = h - a = 225 - 15 = 210 cm.

The height of the compressed zone of concrete in the first approximation

Rice. 3.5 Design diagram of the cross-section of the beam in section 2-2

Because x1? h "p, the section works as rectangular and the required area of ​​the reinforcement

Number of high-strength wire bundles

We take the number of beams n = 17 pcs., Then the area of ​​the reinforcement, the distance from the center of gravity of the reinforcement to the bottom of the beam and the working height

Fa = 17 4.704 = 79.97 cm2, a = 16.5 cm, h0 = 225 - 16.5 = 208.5 cm.

The height of the compressed concrete zone corresponding to the adjusted reinforcement area Fa.

Working condition coefficient

where R0 = 0.3 = 0.3 17000 = 5100 kgf / cm2 (since 0.00015 5100 = 0.765> 0.75, we take 0.00015 5100 = 0.75).

We take m2 = 1.0, then the corrected height of the compressed zone of concrete

xc = m2x2 = 1.0 19.1 = 19.1 cm.

Rice. 3.6 Layout of prestressed reinforcement for 17 beams

Shoulder of a pair of internal forces

z = h0 - 0.5c = 208.5 - 0.5 19.1 = 198.9 cm.

Checking the strength of the section by the bending moment.

Mpr = Faz = 10100 79.97 198.9 = 1606.5 105 kg / cm2 = 1606.5 tf m,

Мпр,> М2 = 1555.1 tf m - the check is in progress.

3.2.3 Analysis of fracture toughness at the stage of manufacture and operation

Checking against the formation of normal cracks during operation.

Rice. 3.7 Schemes for the calculation of the main beam for crack resistance a) at the stage of operation, b) at the stage of manufacture

The calculation is carried out for the greatest bending moment M "from the standard loads at a reduced value of the dynamic coefficient (M" 2 = 1307.3 tf m). It is assumed that in the stage preceding the formation of cracks, concrete and reinforcement retain their elastic properties. Thanks to the pre-stressing, the structure operates at full cross-section.

Tensioning of the reinforcement will be applied to the concrete, considering two stages of work of the structure under load. At the first stage, the structure works with a concrete section, absorbing the forces from the pre-tensioning of the reinforcement in the channels and its own weight.

Determine the geometric characteristics of the concrete section.

The calculated areas of the flange and bottom chord F1 and F3 are determined using AutoCAD 2000

F1 = Fп = 5346 cm2, F3 = Fнп = 3269 cm2.

Estimated width of the slab and bottom chord

bp = 180 cm, bnp = 82 cm.

Estimated height of the shelf and bottom chord

Calculated rib area

F2 = b (h - hp - hnp) = 26 (225 - 29.7 - 39.9) = 4041.3 cm2.

Weakening area by channels

Fo = 17 3.14 2.52 = 333.8 cm2.

Concrete section area

Fb = F1 + F2 + F3 - Fо = 5346 + 4041.3 + 3269 - 333.8 = 12322.5 cm2.

Sbn = F1 (h - 0.5h "p) + 0.5F2 (h - h" p + hnp) + 0.5 (F3 - Fo) hnp == 5346 (225 - 0.5 29.7) + 0 , 5 4041.3 (225 - 29.7 + 39.9) + 0.5 (3269 - 333.8) 39.9 = 1657155.6 cm3.

ubv = h - ybn = 225 - 134.5 = 90.5 cm.

Moment of inertia of a concrete section relative to the neutral axis

Ib = + F1 (ybv - 0.5h "p) 2 + F2 [ub - 0.5 (h - h" p + hnp)] 2 + (F3 - Fo) (ybn - 0.5hnp) 2,

Ib = + 5346 (90.5 - 0.5 29.7) 2 + 4041.32 + (3269 - 333.8) (134.5 - 0.5 39.9) 2 = 79239986.6 cm4.

At the second stage, a load from the weight of the ballast with parts of the track (road surface) and a temporary vertical load act on the beams of the superstructure. At this stage, after the injection of the channels, the reinforcement and the concrete of the structure work together. Geometric characteristics are determined for a reduced section, in which the reinforcement is replaced by concrete of equivalent area. The value of the coefficient of reduction of prestressing reinforcement to concrete for concrete grade M500 and wire reinforcement nн = 6.0.

Reduced (including reinforcement) cross-sectional area

Fп = Fb + nнFа = 12322.5 + 6.0 79.97 = 12802.3 cm2.

The static moment of the reduced section relative to the lower edge of the beam

Sпн = Sbn + nнFаa = 1657155.6 + 6.0 79.97 16.5 = 1665072.6 cm3.

Distance from the bottom edge of the beam to the neutral axis

Distance from the top edge of the beam to the neutral axis

upv = h - ypn = 225 - 130.1 = 94.9 cm.

The moment of inertia of the reduced section relative to the neutral axis

Ip = + F1 (ypv - 0.5h "p) 2 + F2 [upv - 0.5 (h - h" p + hnp)] 2 + (F3 - Fа) (ypn - 0.5hnp) 2 + nnFa ( yпн - a) 2,

Ip = + 5346 (94.9 - 0.5 29.7) 2 + 4041.32 + (3269 - 79.97) (130.1 - 0.5 39.9) 2 + 6.0 79.97 ( 130.1 - 16.5) 2 = 88378591.4 cm4.

Moments of resistance of the lower and upper faces of the concrete and reduced section

Eccentricities of the application of the tensile force of the reinforcement relative to the centers of gravity of the concrete and reduced sections

eb = ubn - a = 134.5 - 16.5 = 118 cm, en = upn - a = 130.1 - 16.5 = 113.6 cm.

The bending moment when calculating the fracture toughness M "can be represented as the sum of the moments from the dead weight M" sv and from the weight of the ballast (road surface) M "wb and from the temporary load M" vr.

M "sv = 22.35 tf m,

M "wb + M" bp = 73.55 tf m.

Taking into account the two stages of operation of the section under load, the expected tensile stresses at the lower edge

These stresses can be extinguished by tensioning the reinforcement with a force N "pr with the transfer of this force to the concrete of the structure. From this condition, we determine the minimum required tensile force of the reinforcement

The stress in the reinforcement from its pre-tension, which persists for the entire period of operation

The stresses ya2 during the tension of the reinforcement should be increased taking into account the inevitable stress losses over time from shrinkage and creep of concrete, relaxation of the reinforcement and from the influence of other factors. Controlled voltages

yak = ya2 + uploss = 1.3ua2 = 1.3 6495.5 = 8444.2 kg / cm2.

0.65 = 0.65 17000 = 11050 kg / cm2

Condition uak? 0.65 is executed.

Checking the crack resistance of the beam at the manufacturing stage.

During the manufacturing stage, the prestressing force and its own weight act on the structure. At this stage, we check the compressive stresses in the extreme fiber of the lower belt in section 2-2

Rt = 225 kg / cm2.

Condition ubn? RT is executed.

When prestressing, tensile stresses can occur in the upper region of the beam. Stresses in the upper fiber of the section

Rr.p. = 17.5 kg / cm2.

Condition ubn? Rp.r. is carried out, additional reinforcement of the upper zone of the beam is not required.

Checking the stresses in the reinforcement during the initial period of operation.

The prestressing stresses of the reinforcement are summed up with the stresses from the action of the operational loads on the structure. The stresses in the reinforcement of the extreme lower row are checked

Rн2 = 10100 kg / cm2.

Condition ubn? Rн2 is executed.

3.2.4 Design for shear and principal stresses

The calculation is carried out for the forces M "and Q" from the standard loads and the effect of the pretensioning force N "pr. It is assumed that during the operation stage the structure works elastically and with a full section. When performing checks, the magnitude of the acting shear and principal stresses is limited. Checking the shear stresses.

Shear stresses are determined in the web of the main beam in the sections above the support and in the middle of the span at three points along the height of the section: in the places where the slab and the lower chord adjoin the web and on the neutral axis.

Rsk = 45 kgf / cm2.

Section 2-2 (Q "2 = 41.3 tf, b = 26 cm, Ip = 88378591.4 cm4):

y1 = upv - 0.5h "n = 94.9 - 0.5 29.7 = 80.1 cm,

S1 = F1y1 = 5346 80.1 = 428159.6 cm3,

y2 = (upv -h "n) 0.5 = (94.9 -29.7) 0.5 = 32.6 cm,

S2 = b (ypv - h "p) y2 + F1y1 = 26 (94.9 -29.7) 32.6 + 5346 80.1 = 483490.5 cm3,

y3 = upn -0.5hpn = 130.1- 0.5 39.9 = 110.1 cm,

S3 = F3y3 = 3269 110.1 = 360006.4 cm3,

Section 0-0 (Q0 = 212.1 tf, b = 82 cm):

calculated flange area F1 determined by AutoCAD 2000

F1 = Fп = 7026 cm2,

calculated slab width

calculated shelf height

calculated rib area

F2 = b (h - h "p) = 26 (225 - 39.0) = 15249.3 cm2,

channel weakening area

Fo = 17 3.14 2.52 = 333.8 cm2,

concrete section area

Fb = F1 + F2 - Fo = 7026 + 15249.3 - 333.8 = 21941.5 cm2,

Rice. 3.8 Design model of the cross-section of the beam in the section 0-0

Static moment of a concrete section relative to the bottom edge of the beam

Sbn = F1 (h - 0.5h "p) + 0.5 (F2 - Fo) (h - h" p) = 5346 (225 - 0.5 39.0) + 0.5 (15249.3 - 333.8 ) (225 - 39.0) = 2830616.2 cm3,

reduced (including reinforcement) cross-sectional area

Fп = Fb + nнFа = 21941.5 + 6.0 79.97 = 22421.3 cm2,

static moment of the reduced section relative to the lower edge of the beam

Sпн = Sbn + nнFаa = 2830616.2 + 6.0 79.97 16.5 = 2838533.3 cm3,

distance from the bottom edge of the beam to the neutral axis

distance from the top edge of the beam to the neutral axis

upv = h - ypn = 225 - 126.6 = 98.4 cm,

moment of inertia of the reduced section relative to the neutral axis

Iп = + F1 (yпв - 0.5h "п) 2 + (F2 - Fа) 2 + nнFа (yпн - a) 2,

Ip = + 7026 (98.4 - 0.5 39.0) 2 + (15249.3 - 79.97) [126.6 - 0.5 (225 - 39.0)] 2 + 6.0 79.97 (126.6 - 16, 5) = 111518701.2 cm4,

eccentricity of the application of the tensile force of the reinforcement relative to the center of gravity of the reduced sections

en = upn - a = 126.6-16.5 = 110.1cm;

shear force of deflected beams of prestressed reinforcement

Q "but = уа2Уfоsinbi = 6495.5 3 4.704 sin14? = 22175.6 kgf;

y1 = upv - 0.5h "n = 98.4 - 0.5 39.0 = 78.9 cm,

S1 = F1y1 = 7026 78.9 = 554351.4 cm3,

y2 = (upv -h "p) 0.5 = (98.4 - 39.0) 0.5 = 29.7 cm,

S2 = b (ypv - h "p) y2 + F1y1 = 82 (98.4 -39.0) 29.7 + 7026 78.9 = 699014.2 cm3,

y3 = 106.7 cm,

S3 = F3y3 = 3269 106.7 = 348802.3 cm3,

All checks are carried out.

Checking the main stresses.

The main tensile and main compressive stresses are determined in the web of the main beam in the sections above the support and in the middle of the span at three points along the height of the section: at the places where the slab and the lower chord adjoin the wall and on the neutral axis.

Reinforcement of the beam with tensioned clamps is not performed, therefore

Re = 205 kgf / cm2, Rrp = 17.5 kgf / cm2.

Section 2-2 (M "2 = 1307.3 tf m):

normal surveys in a beam section arising from prestressing force and bending moment

y1 = upv - h "n = 94.9 - 29.7 = 65.2 cm,

at point 2 y2 = 0,

y3 = - (upn - hnp) = - (130.1 - 39.9) = 90.2 cm,

Section 0-0 (M "2 = 0): normal surveys in the section of a beam arising from prestressing force and bending moment

y1 = upv - h "n = 98.4 - 39.0 = 59.4 cm,

at point 2 y2 = 0,

at point 3 y3 = - 90.2 cm,

In all cases, the conditions are ugs? Re and eel? 0.8о1Rрп are fulfilled.

3.2.5 Shear strength design

The calculation is performed in a section formed by an inclined crack. The transverse force is perceived by deflected reinforcement beams, clamps and concrete in the compressed section zone.

The clamps are taken from reinforcement class A-III d = 8 cm (= 4000 kgf / cm2, Ra = 3100 kgf / cm2, f = 0.503 cm).

Section 2-2 (Q2 = 49.6 tf, b = 26 cm):

Qxb = Q2 - Qno = 49.6 tf;

pitch of conventional clamps

according to the design requirements, we take the step of the clamps ah = 20 cm.

Section 0-0 (Q0 = 251.9 tf, b = 82 cm):

part of the shear force perceived by deflected beams

Qno = RnoUfosinbi = 0.7Rn2Ufosinbi = 0.7 10100 3 4.704 sin14? = 22444 kgf

lateral force perceived in an inclined section by clamps and concrete in a compressed zone

Qхb = Q0 - Qno = 251900 - 22444 = 229456 kgf;

force in clamps per unit of beam length

pitch of conventional clamps

according to the design requirements, we take the step of the clamps ah = 10 cm.

bridge span beam tension

Literature

1. Calculation of reinforced concrete bridges. Ed. K.K. Jacobson. - M .: Transport, 1977.

2. Reinforced concrete bridges. Development of options: guidelines to course and diploma design. Ch. 1.2 - L .: LIIZHT, 1966

3. E.I. Ivanov, E. S. Karapetov, E. D. Maksarev Calculation of girder reinforced concrete bridges: guidelines for course design. - L .: LIIZHT, 1983.

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Lecture plan

4.1. Scope, basic systems, materials

4.2. Structures of span structures of girder bridges

4.3. Continuous beam bridges

4.4. General information about frame and arched bridges

4.5. The main provisions for the design of reinforced concrete beam-split spans

4.1. Applications, basic systems and materials

On the railways of Russia, mainly small and medium-sized reinforced concrete bridges are used.

According to their design features, the spans of reinforced concrete bridges are divided into two types: with non-tensioned reinforcement and with prestressed reinforcement of the main beams.

They are single-track and double-track, but preference is given to superstructures with one branch of the railway track.

To the main systems reinforced concrete bridges include beam (split, continuous and cantilever), frame, arched.

Beamsplit reinforced concrete spans have received the most widespread use (Fig. 4.1, a).

Rice. 4.1. The main systems of reinforced concrete bridges: a- split beam; b- continuous beam; v- cantilever beam; G- frame; d- arched

They are used primarily for small to medium sized bridges. Beam continuous structures are used to cover large spans (Fig.4.1, b). In terms of material consumption, they are more economical in comparison with simple split systems, but they have limitations in application due to their sensitivity to uneven settlement of supports, shrinkage and creep of concrete, as well as temperature deformations. Framed reinforced concrete bridge systems are characterized by a rigid connection of the girder and the rack, working together (Fig. 4.1, G). Their advantage over simple beam-split systems lies in increased structural rigidity and lower material consumption, but at the same time they have the same disadvantages as continuous spans. Arched Spans are used to cover large and giant spans. Their advantage over split spans lies in the fact that arches, which work mainly in compression, to the greatest extent provide the strength properties of reinforced concrete (Fig. 4.1, d). Arched spacer and non-spacer bridges are used, as well as hingeless and hinged systems. Arched bridges are durable, but very labor intensive and expensive.

Combined reinforced concrete bridges are used, in which the work of two or more systems is combined. These include bridges with arched spans with a ride in the middle, as well as cable-stayed and hanging spans (Fig. 4.2).

Such bridges are distinguished by their architectural merits and more economic performance and, as a rule, are used to bridge large, giant and super-giant spans. Cable-stayed and suspended spans are mainly used in the system of road and city bridges.

Reinforced concrete Is a complex building material consisting of concrete and steel reinforcement (1–4%), working together under load. When distributing functions between concrete and reinforcement, a condition is envisaged under which the concrete ensures the operation of structures mainly in compressed zones, and steel reinforcement in tension zones.

The advantages of reinforced concrete bridges include high strength, durability, fire resistance, ability to resist when exposed to natural and climatic factors, and low operating costs.

Concrete. For elements of reinforced concrete bridges, structural heavy concrete with an average density of 2200-2500 kg / m 3 is used.

The main characteristic that determines the strength properties includes compressive strength class of concrete. The compressive strength class of concrete is expressed by the normative resistance to axial compression of cubes 15 - 15 - 15 cm in size with a security of 0.95, measured in megapascals. Dependence between concrete class V the compressive strength and concrete strength determined on cubes are expressed by the dependence

, (4.1)

where
- coefficient of variation of concrete strength, which, according to the normative documents for heavy concrete, is taken = 0,135;
- standard deviation of concrete strength values ​​in a series of test specimens; Is the average value of concrete strength in a series of samples.

For the structures of reinforced concrete bridges, concrete of classes B20 is used; B22.5; B25; B27.5; B30; B40; B45; B50; B55; B60.

Concrete is an elastoplastic material in which elastic and plastic deformations develop simultaneously under the action of a load. The ratio of stress in concrete to elastic relative deformations determines the elastic properties of the material, characterized by modulus of elasticity of concrete
... The modulus of elasticity of concrete has the same value in compression and tension and depends on the class of concrete in terms of strength and hardening conditions, it is determined according to SNiP 2.05.03-84 * depending on the class of concrete.

Requirements are imposed on the concrete of bridge structures frost resistance depending on the climatic conditions of construction and operation. Concrete grade for frost resistance determined according to SNiP 2.05.03-84 *.

Concrete grade on water resistance, characterizing the density and mobility of the concrete mixture, is determined according to SNiP 2.05.03-84 *.

During the construction, repair or reconstruction of bridges, significant characteristics include strength gain rate concrete. According to ordinary concrete, it reaches 50% strength after 3 days at a temperature of plus 20 ° C, and when the concrete mixture is heated and steamed, it can gain up to 80% strength after 2 days.

Armature is an integral part of reinforced concrete. The requirements for reinforcement are that it must reliably ensure joint work with concrete at all stages of operation of bridge structures, be used up to the physical or conditional yield strength when their bearing capacity is exhausted, and also comply with the mechanization conditions during installation work.

The reinforcement of the elements of reinforced concrete bridges is subdivided into working and structural. Under working they understand reinforcement, the cross-sectional area of ​​which is determined by calculating the action of external loads. TO constructive include mounting and distribution fittings installed without calculation for structural or technological reasons. Mounting reinforcement provides the rigidity of the reinforcement cage. Distribution reinforcement is designed for a more even distribution of concentrated forces in the bars of the working reinforcement. Structural reinforcement is also installed for partial perception of forces that are not taken into account by the calculation from shrinkage and creep of concrete, temperature stresses, local stresses from concentrated forces, random stresses arising during the manufacture, transportation and installation of structures.

Reinforcement is subdivided into hot-rolled rod, cold-drawn wire and thermally hardened smooth and periodic profile, non-stressed and stressed.

Reinforcing steel is characterized by class and brand. The reinforcement class determines the strength properties of the steel. The grade of low-alloy steel indicates its chemical composition, and carbon steel indicates information about the degree of deoxidation, group and category of guarantee.

Non-tensioned bar reinforcement is used in classes A-I, A-II, A with -II, A-III with a diameter of 6 to 40 mm. Strain reinforcement is used from wire with a diameter of 3-5 mm of class B-II in the form of bundles, as well as high-strength bar reinforcement of periodic profile of classes A-IV, A-V, A-VI.

The main strength characteristic of reinforcing steel is the physical or conditional yield strength. The physical yield strength is typical for steels of the classes used for non-tensioned reinforcement, and the conditional one is for high-strength bar and high-strength prestressing reinforcement. High-strength reinforcing steel is characterized by a conventional yield point, which is taken as a stress with a residual relative deformation of 0.2%. The main indicator of the strength of hard steels is the tensile strength.

Stress relaxation is characteristic of prestressed high-strength reinforcement. It depends on the strength and chemical composition of the steel, manufacturing technology, temperature, rebar tension and other factors. Relaxation of stresses proceeds unevenly: it is most intense in the first hours, and then the process gradually dies out.